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Let $a,b,c>0$. Prove $$\sqrt{\frac{a}{b+3c}}+\sqrt{\frac{b}{c+3a}}+\sqrt{\frac{c}{a+3b}}\ge \frac{3}{2}$$


$A=\sqrt{\frac{a}{b+3c}}+\sqrt{\frac{b}{c+3a}}+\sqrt{\frac{c}{a+3b}}$

Holder: $A^2\cdot Σ_{cyc}\left(a^2\left(b+3c\right)\right)\ge \left(a+b+c\right)^3$

So we need to prove $4\left(a+b+c\right)^3\ge 9Σ_{cyc}\left(a^2\left(b+3c\right)\right)$

$\Leftrightarrow 4a^3+4b^3+4c^3+3a^2b+3b^2c+3c^2a+24abc\ge 15ab^2+15bc^2+15a^2c$

Then i don't know how to solve it. Help me !

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  • $\begingroup$ Is any condition else given? $\endgroup$ – Dr. Sonnhard Graubner Apr 20 '18 at 13:17
  • $\begingroup$ @Dr. Sonnhard Graubner: no $\endgroup$ – Word Shallow Apr 20 '18 at 13:33
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Your trying gets a wrong inequality.

Try $c=0$ and $a=b=1$.

Let $\frac{a}{b+3c}=\frac{x^2}{4}$, $\frac{b}{c+3a}=\frac{y}{4}$ and $\frac{c}{a+3b}=\frac{z^2}{4}$, where $x$, $y$ and $z$ are non-negative numbers.

Hence, the system $$\begin{array}{l}4a-x^2b-3x^2c=0\\-3y^2a+4b-y^2c=0\\ -z^2a-3z^2b+4c=0\\ \end{array}$$ has infinitely many solutions $(a,b,c)$,

Hence $$ \|(\begin{array}{ccc}4& -x^2 & -3x^2\\ -3y^2 & 4 & -y^2\\ -z^2 & -3z^2 & 4\end{array})\| = 0 ,$$ which gives $$3(x^2y^2+x^2z^2+y^2z^2)+7x^2y^2z^2=16$$ and we need to prove that $$x+y+z\geq3.$$ Let $x+y+z<3$, $x=kp$, $y=kq$ and $z=kr$ such that $k>0$ and $p+q+r=3$.

Hence, $k(p+q+r)<3$, which gives $0<k<1$.

Thus, $$16=3(x^2y^2+x^2z^2+y^2z^2)+7x^2y^2z^2=3k^4(p^2q^2+p^2r^2+q^2r^2)+7k^6p^2q^2r^2<$$ $$<3(p^2q^2+p^2r^2+q^2r^2)+7p^2q^2r^2.$$ But it's a contradiction because we'll prove now that $$16\geq3(p^2q^2+p^2r^2+q^2r^2)+7p^2q^2r^2$$ for all non-negatives $p$, $q$ and $r$ such that $p+q+r=3$.

Indeed, let $p+q+r=3u$, $pq+pr+qr=3v^2$ and $pqr=w^3$.

Hence, $16\geq3(p^2q^2+p^2r^2+q^2r^2)+7p^2q^2r^2\Leftrightarrow f(w^3)\leq0$, where $f$ is a convex function.

Hence, by $uvw$ it remains to check two cases.

  1. $w^3=0$.

Let $r=0$.

We need to prove that $$16\geq3p^2q^2,$$ which is true because $$p^2q^2\leq\frac{(p+q)^4}{16}=\frac{81}{16}<\frac{16}{3}.$$

  1. $q=r$.

After homogenization we can assume $q=r=1$ and it remains to prove that $$\frac{16(p+2)^6}{729}\geq7p^2+\frac{(2p^2+1)(p+2)^2}{3}$$ or $(p-1)^2(8p^4+112p^3+453p^2+1102p+26)\geq0$, which is obvious.

Done!

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  • $\begingroup$ But $c>0$ why do you choose $c=0$ ? $\endgroup$ – Word Shallow Jun 9 '18 at 13:00
  • $\begingroup$ Try $c=0.0001$ and $a=b=1$. $\endgroup$ – Michael Rozenberg Jun 9 '18 at 14:13

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